Math 447
Final Exam
Spring 2009
Principles of Analysis II
Instructions Solve four of the following six problems. Please write your
solutions on your own paper. Explain your reasoning in complete sentences.
1. Let f be a function of bounded variation on the interval [0, 1]. Suppose there is a positive number δ such that |f (x)| ≥ δ for every x (in
other words, the function f is bounded away from 0). Show that the
reciprocal function 1/f is a function of bounded variation.
Solution. Suppose 0 = x0 < x1 < · · · < xn = 1. These points xk
define a partition P of the interval [0, 1]. The quantity V (1/f, P ), the
variation of the function 1/f with respect to the partition P , equals
n X
1
1
f (xk ) − f (xk−1 ) .
k=1
Observe that
1
1 |f (xk ) − f (xk−1 )|
1
f (xk ) − f (xk−1 ) = |f (xk )f (xk−1 )| ≤ δ 2 |f (xk ) − f (xk−1 )|
for each k. Therefore V (1/f, P ) ≤ δ −2 V (f, P ) ≤ δ −2 V01 (f ), where
V01 (f ) denotes the total variation of f . Taking the supremum over all
partitions P shows that the function 1/f has bounded variation, and
moreover the total variation of this function does not exceed δ −2 V01 (f ).
2. Give a concrete example of a uniformly convergent sequence (fn ) of
functions of bounded variation on the interval [0, 1] such that the limit
function does not have bounded variation.
Solution. You saw in class the standard example of a continuous
function f that does not have bounded variation: namely,
(
x sin(1/x), if x 6= 0,
f (x) =
0,
if x = 0.
To make use of this example, define fn as follows:
(
x sin(1/x), if x > 1/n,
fn (x) =
0,
if x ≤ 1/n.
May 11, 2009
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Dr. Boas
Math 447
Final Exam
Spring 2009
Principles of Analysis II
The function fn has bounded variation because the graph consists of
finitely many bounded, monotonic pieces. (The number of pieces grows
with n, however.) The sequence (fn ) converges uniformly to f because
sup |fn (x) − f (x)| ≤ 1/n.
0≤x≤1
3. If α is a nondecreasing function
R π on the closed interval [−π, π], is it
necessarily true that limn→∞ −π cos(nx) dα(x) = 0? (In other words,
does the Riemann–Lebesgue lemma carry over to the setting of the
Stieltjes integral?) Give either a proof or a counterexample.
Solution. For a counterexample, consider the step function α such that
α(x) = 0 if x < 0, and α(x) = 1 if x ≥ 0. You know from class and from
the homework exercises that StieltjesRintegration against this α acts like
π
a “delta function”: in other words, −π cos(nx) dα(x) = cos(0) = 1 for
Rπ
every n. Hence limn→∞ −π cos(nx) dα(x) 6= 0.
Remark If α isRdifferentiable and hasR a derivative that is Riemann
π
π
integrable, then −π cos(nx) dα(x) = −π cos(nx)α0 (x) dx [by Theorem 14.18 on page 232]; in this case, the limit of the integral is equal to 0
by the usual Riemann–Lebesgue lemma [equation (15.4) on page 248
or Theorem 19.17 on page 353].
4. Let f be a bounded function that is Riemann–Stieltjes integrable with
respect to the increasing function α on the interval [0, 1]. Prove that f is
Riemann–Stieltjes Rintegrable with respect to αR2 on the same interval.
1
1
In other words, if 0 f dα exists, then so does 0 f d(α2 ).
Remark added May 12 The solution below assumes that the function α2 is increasing, which need not be the case. [The function α(x)
might be x − 1/2, for example.] There are a couple of ways to reduce
to the situation in which α2 is increasing.
1. Either α is nonnegative (in which case α2 is increasing), or α is
nonpositive (in which case α2 is decreasing, and −α2 is increasing), or
there is a point c where α changes sign from negative to positive (in
which case α2 is decreasing on the interval [0, c] and increasing on the
May 11, 2009
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Dr. Boas
Math 447
Final Exam
Spring 2009
Principles of Analysis II
interval [c, 1]). In the first case, the argument
R 1 below 2applies. In the
second case, the argument below shows that 0 f d(−α ) exists, and it
R1
R1
follows from the definitions that 0 f d(−α2 ) = − 0 f d(α2R). In the
c
third case, use the preceding two cases to deduce that both 0 f d(α2 )
R1
R
1
and c f d(α2 ) exist; hence 0 f d(α2 ) exists.
2. Alternatively, observe that there is a constant k for which the function α + k is positive (any value
than |α(0)| will do). The
R 1 of k greater
2
argument below implies that 0 f d((α + k) ) exists. It is routine to see
R1
R1
that existence of 0 f dα implies existence of 0 f d(2kα + k 2 ) [which
R1
R1
R1
equals 2k 0 f dα], and the difference 0 f d((α+k)2 )− 0 f d(2kα+k 2 )
R1
equals 0 f d(α2 ).
Solution. Fix an arbitrary positive ε. The integrability of f with
respect to α implies (by Riemann’s condition) that there is a partition P
such that the upper sum Uα (f, P ) and the lower sum Lα (f, P ) differ
by less than ε. In other words, there is a subdivision of the interval
[0, 1] into n subintervals [xk−1 , xk ] such that if Mk and mk denote the
supremum and the infimum of f on [xk−1 , xk ], then
n
X
(Mk − mk )[α(xk ) − α(xk−1 )] = Uα (f, P ) − Lα (f, P ) < ε.
k=1
The goal is to estimate the difference between upper and lower sums
with respect to α2 . The function α is increasing, so α(xk )2 −α(xk−1 )2 =
[α(xk ) − α(xk−1 )][α(xk ) + α(xk−1 )] ≤ 2α(1)[α(xk ) − α(xk−1 )]. Therefore
n
X
Uα2 (f, P ) − Lα2 (f, P ) =
(Mk − mk )[α(xk )2 − α(xk−1 )2 ]
k=1
≤ 2α(1)[Uα (f, P ) − Lα (f, P )] ≤ 2εα(1).
Since ε is arbitrary, Riemann’s condition implies that f is integrable
with respect to α2 .
5. Determine the Fourier series of the odd function on the interval [−π, π]
that is equal to 1 on the interval (0, π), and use the result to compute
∞
X
1
.
the value of the numerical series
(2k + 1)2
k=0
May 11, 2009
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Dr. Boas
Math 447
Final Exam
Spring 2009
Principles of Analysis II
Solution.
P The Fourier series of an odd function is a sine series of the
form ∞
n=1 bn sin(nx). For the specified function,
π
Z
2 π
2 − cos(nx)
bn =
1 · sin(nx) dx =
π 0
π
n
0
(
4
, if n is odd,
2
[− cos(nπ) + cos(0)] = nπ
=
nπ
0,
if n is even.
Writing an odd integer n in the form 2k + 1 shows that the Fourier
series has the form
∞
X
k=0
4
sin[(2k + 1)x].
(2k + 1)π
According to Parseval’s equation, the sum of the squares of the Fourier
coefficients equals 1/π times the integral of the square of the function
over the interval [−π, π]. Thus
∞
1
2
16 X
=
2
2
π k=0 (2k + 1)
π
Z
π
2
1 dx = 2,
0
so
∞
X
k=0
1
π2
=
.
(2k + 1)2
8
Remark You summed this series using a different Fourier series in
Assignment 8.
6. Suppose f ∈ L2 [−π, π]. Then sn (f ), the nth partial sum of the Fourier
series of f , has the property that limn→∞ ksn (f )−f k2 = 0 (according to
the Riesz–Fischer theorem). Use this result to prove that the Cesàro
sum σn (f ), which is the average [s0 (f ) + · · · + sn−1 (f )]/n, has the
corresponding property that limn→∞ kσn (f ) − f k2 = 0.
Solution. Fix an arbitrary positive ε. By the Riesz–Fischer theorem,
there is a positive integer N such that ksk (f ) − f k2 < ε/2 when k ≥ N .
May 11, 2009
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Dr. Boas
Math 447
Final Exam
Spring 2009
Principles of Analysis II
Thus when n > N , the definition of σn (f ) implies that
[s0 (f ) − f ] + · · · + [sn−1 (f ) − f ] kσn (f ) − f k2 = n
2
N
−1
n−1
X
X
1
1
ksk (f ) − f k2 +
ksk (f ) − f k2
≤
n k=0
n k=N
N −1
ε 1X
ksk (f ) − f k2 .
< +
2 n k=0
Since N is fixed (dependent on ε but not on n), the second term in
the third line of the displayed formula will be less than ε/2 when n is
sufficiently large. Hence kσn (f ) − f k2 < ε when n is sufficiently large.
Therefore limn→∞ kσn (f ) − f k2 = 0.
Bonus problem For extra credit, prove either the Riesz representation
theorem characterizing the dual space of C[0, 1] or Jordan’s decomposition
theorem for functions of bounded variation.
Solution. We did these proofs in class, and there are proofs in the book
too (pages 237–239 and 207).
May 11, 2009
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Dr. Boas